The values of two resistors are R 1 = 6 ± 0.3 k Ω and R 2 = 10 ± 0.2 k Ω . The percentage error in the equivalent resistance when they are connected in parallel is

Step-by-step solution

R 1 = 6 ± 0.3 k Ω , R 2 = 10 ± 0.2 k Ω R p a r a l l e l = R 1 R 2 R 1 + R 2 Let R 1 + R 2 = x ⇒ R P = R 1 R 2 x Taking log of both sides l n R P = l n R 1 + ln ⁡ R 2 - ln ⁡ x Differentiating, ∆ R P R P = ∆ R 1 R 1 + ∆ R 2 R 2 + - ∆ x x ∆ x m e a n = 0.3 + 0.2 2 = 0.25 Ω R m e a n = 6 + 10 2 = 8 Ω ∴ x = 6 + 10 2 = 8 Ω ⇒ ∆ x x = 0.25 8 ∴ T o t a l e r r o r = 0.3 6 + 0.2 10 + 0.25 8 = 0.05 + 0.02 + 0.03125 = 0.10125 ∴ ∆ R P R P = 10.125 %